/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Find the angle between \(\mathbf... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 22

Find the angle between \(\mathbf{v}\) and \(\mathbf{w} .\) Round to the nearest tenth of a degree. $$\mathbf{v}=3 \mathbf{j}, \quad \mathbf{w}=4 \mathbf{i}+5 \mathbf{j}$$

Short Answer

Expert verified
The angle between the vectors \( \mathbf{v} \) and \( \mathbf{w} \) is approximately \( 35.5^\circ \) .

Step by step solution

01

Understand the Concept of Dot Product

Firstly, one must understand the concept of dot product (or scalar product) of two vectors, which involves both the magnitude and direction of vectors. This will be useful for the next step.
02

Apply the Dot Product Formula

Given the formula for dot product which is equal to the product of the magnitudes of the two vectors and the cosine of the angle between them. Written as \( \mathbf{v} \cdot \mathbf{w} = ||\mathbf{v}|| ||\mathbf{w}|| cos(\theta) \) where \( \theta \) is the angle between the two vectors, \( \mathbf{v} \) and \( \mathbf{w} \), and \( ||\mathbf{v}||, ||\mathbf{w}|| \) are the magnitudes of the vectors. Using this formula, we can rearrange for \(\cos(\theta)\) where \(\cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}|| ||\mathbf{w}||}\) .
03

Determine the Dot Product

The dot product \( \mathbf{v} \cdot \mathbf{w} \) is computed by multiplying corresponding components of the two vectors together, and then taking their sum. In this case, \( \mathbf{v} \cdot \mathbf{w} \) is the sum of the product of the two i components and the two j components: \( \mathbf{v} \cdot \mathbf{w} = 0*4 + 3*5 = 15 \) .
04

Determine the Magnitude of Vectors

The magnitude of a vector in component form is given by the Pythagorean theorem \( ||\mathbf{v}|| = \sqrt{i^2 + j^2} \). Thus, the magnitude of \( \mathbf{v} \) is \( ||\mathbf{v}|| = \sqrt{0^2 + 3^2} = 3 \), and the magnitude of \( \mathbf{w} \) is \( ||\mathbf{w}|| = \sqrt{4^2 + 5^2} = \sqrt{41} \) .
05

Compute Calculation for Cosine of the Angle

Substituting the above values into the formula from Step 2, \(\cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}|| ||\mathbf{w}||} = \frac{15}{3 \sqrt{41}} = \frac{5}{\sqrt{41}}\) .
06

Determine the Angle

Now to find the angle, \( \theta \), take the inverse cosine (or arccos) of the value calculated in the previous step, \( \theta = arccos \left( \frac{5}{\sqrt{41}} \right) \). To present the result in degrees rather than in radians, the value should be converted with the appropriate conversion factor. According to most calculators, this results in approximately \( \theta = 35.5^\circ \) .

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain how to find the dot product of two vectors.

Graph: \(\quad f(x)=\frac{4 x-4}{x-2}\)

Two buildings of equal height are 800 feet apart. An observer on the street between the buildings measures the angles of elevation to the tops of the buildings as \(27^{\circ}\) and \(41^{\circ} .\) How high, to the nearest foot, are the buildings?

Find \(\text {pro}_{\mathbf{w}} \mathbf{V}\) Then decompose v into two vectors, \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2},\) where \(\mathbf{v}_{1}\) is parallel to \(\mathbf{w}\) and \(\mathbf{v}_{2}\) is orthogonal to \(\mathbf{w}.\) $$\mathbf{v}=2 \mathbf{i}+\mathbf{j}, \quad \mathbf{w}=6 \mathbf{i}+3 \mathbf{j}$$

Use the dot product to determine whether v and w are orthogonal. $$\mathbf{v}=8 \mathbf{i}-4 \mathbf{j}, \quad \mathbf{w}=-6 \mathbf{i}-12 \mathbf{j}$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.